We know that if (x+a) is a zero of a function f(x) then f(−a)=0.
Given is f(x)=x3−3x2−15x+125
Let us choose one factor as (x+5). This is trial and error, notice that there is 125 as the last term therefore this is a possible factor.
Now f(−5)=(−5)3−3(−5)2−15(−5)+125
or f(−5)=−125−75+75+125=0
Hence (x+5) is a factor.
Writing the given equation as a multiplication of the factor and a quadratic in general form
(x+5)(ax2+bx+c)=x3−3x2−15x+125
ax3+bx2+cx+5ax2+5bx+5c=x3−3x2−15x+125
Put like terms together
ax3+bx2+5ax2+cx+5bx+5c=x3−3x2−15x+125
ax3+x2(b+5a)+x(c+5b)+5c=x3−3x2−15x+125
Compare coefficients of like terms
a=1
b+5a=−3 , Inserting value of a gives us b=−8
c+5b=−15, Inserting value of b gives us c=25
5c=125 , Gives us again c=25
Now rewrite polynomial as
f(x)=(x+5)(x2−8x+25)
To find remaining two zeros using quadratic formula we obtain
x=−b±√b2−4ac2a
x=−(−8)±√(−8)2−4×1×252×1
x=8±√64−1002
x=8±√−362
x=8±6i2
x=4±3i
